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I have the following integral \begin{align*} \int_{-\infty}^\infty f(t) q(t+ax) dt \end{align*}

where a is some constant.

This integral look a lot like convolution (or correlation). My question is can it be re-write as an convolution of $ f(???)* g(???)$? What are the arguments of the two functions? For example if the integral was $\int_{-\infty}^\infty f(t) q(at+ax) dt$ then we can re-write as \begin{align*} \int_{-\infty}^\infty f(t) q(at+ax) dt= f(x)*q(ax) \end{align*}

This is what I tried I did some integral manipulation and I got this identity \begin{align*} \int f(t) q(t+ax) dt=\int f(t-ax) q(t) dt \end{align*}

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A convolution integral is of the form

$$\int f(t) g(x-t) \, dt;$$

the important point is the structure of the arguments:

$$x-t \qquad \text{and} \qquad t$$

In order to write the given integral as a convolution integral, we define $\tilde{q}(t) := q(-t)$. Then

$$\int_{-\infty}^{\infty} f(t) q(t+ax) \, dt = \int_{-\infty}^{\infty} f(t) \tilde{q}(-ax-t) \, dt = (f \ast \tilde{q})(-ax).$$

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  • $\begingroup$ So is this a convolution of $f(x)$ with $\bar{q}(-ax)=q(ax)$??? $\endgroup$ – Boby Dec 28 '14 at 15:47
  • $\begingroup$ @Boby Note that $(f*\tilde{q})(-ax) \neq (f*q)(ax)$. The first expression means: Calculate the convolution of $f$ with $\tilde{q}$ and then evaluate this function at $-ax$. $\endgroup$ – saz Dec 28 '14 at 16:13
  • $\begingroup$ You see I want apply fourier transform and use the fact that convolution is a product. I have to know what the time domain functions $\endgroup$ – Boby Dec 28 '14 at 16:15
  • $\begingroup$ @Boby So you have to calculate the Fourier transform of $\tilde{q}$. Where are you stuck? $\endgroup$ – saz Dec 28 '14 at 16:34
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    $\begingroup$ @Boby It is gonna be $\frac{1}{|a|}F(\frac{\omega}{-a})\tilde{Q}(\frac{\omega}{-a})$ $\endgroup$ – Sina Dec 29 '14 at 19:07

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